# Determining a consistent estimator/asymptotic relative efficiency

Question:

Let $X_1,\ldots,X_n$ be i.i.d. as $N(0,\sigma^2)$.

a) Show that $\delta_1 = k \sum_{i=1}^n |X_i|/n$ is a consistent estimator of $\sigma$ if and only if $k = \sqrt{\pi/2}$.

b) Determine the asymptotic relative efficiency of $\delta_1$ with respect to $\delta_2 = \sqrt{\sum X_i^2/n}$

Attempt:

a)

i) Given $k = \sqrt{\pi/2}$. We must find a variance that converges to 0.

Calculating the variance of $\delta_1$:

\begin{align} & = E\left[ \left(\sqrt{\frac\pi2} \sum_{i=1}^n |X_i|/n\right)-\mu\right]^2 \\[8pt] & = \frac \pi 2 E\left[ \left(\sum_{i=1}^n |X_i|/n\right)-\mu\right]^2 \end{align}

I'm stuck here, however, this should come to:

$$\frac{\pi}{2}\frac{1}{n}\left[\sigma^2 + \frac{2}{\pi}\sigma^2\right]$$

which converges to zero.

ii) Since $\delta_1$ converges in probability to $\sigma$, and since k must be unique, k must equal $\sqrt{\pi/2}$

b) I know this involves the Central Limit Theorem; however, I'm not able to piece the whole thing together.

I know that

$$E(|X_i|) = \int_{-\infty}^\infty |x|\frac{1}{\sqrt{2\pi\sigma}}\exp\left(\frac{-x}{2\sigma^2}\right)\,dx = \sqrt{\frac{2}{\pi}}\sigma$$

and $\operatorname{Var}(|X_i|) = \sigma^2 - \frac{2}{\pi}\sigma^2$

Also,

$$\sqrt{n}(\delta_1 - \sigma)\xrightarrow[]{L} N\left(0,\left(\frac{\pi}{2}-1\right)\sigma^2\right)$$

Not sure how to proceed.

• Let me suggest to reread carefully (a) since what you propose afterwards is not even related to it. Note that this is $\delta_1$, not $k$, which is a candidate to be an estimator of $\sigma$. And that $k$ is just a constant to be chosen, not an estimator of anything. – Did Oct 11 '14 at 10:42
• Yes that was a mistake. I have updated the question. – statsguyz Oct 14 '14 at 1:51

Please consider: If $X_1,\ldots,X_n$ are identically and independently distributed, then by the strong law of large numbers:

$\delta_1 \rightarrow kE|X|$ with $E|X|={\sigma}\sqrt{2/\pi}.$ So $\delta_1$ will converge to $\sigma$ iff $k = \sqrt{\pi/2}$.

Now we'll need $Var(|X|)=E(X^2)-E^2(|X|)=\sigma^2(1-2/\pi).$

The variance of $\delta_1$ can be computed directly since $\delta_1$ is a linear combination of independent random variables. We obtain:

$$Var(\delta_1)=k^2\frac{Var(|X|)}n=\frac{\sigma^2}n\left(\frac{\pi}2-1\right).$$

Now, from the central limit theorem:

$\sqrt{n}(\delta_1-\sigma)\to N(0,\frac{\sigma^2(\pi-2)}{2})$

Consider $\sum X_i^2/n.$ Its expectation is clearly $\sigma^2$ and $Var(X^2)=E(X^4)-E^2(X^2)=3\sigma^4-\sigma^4=2\sigma^4$

Apply the CLT to $\sum X_i^2/n.$

$\sqrt{n}(\sum X_i^2/n-\sigma^2)\to N(0,2\sigma^4)$

Now we will use the delta method to find the appropriate CLT for $\delta_2.$ Let $g(x)=\sqrt{x}.$ The delta method allows us to conclude:

$$\sqrt{n}(\delta_2-g(\sigma^2))\to N(0,2\sigma^4[g'(\sigma^2)]^2)$$

$$\sqrt{n}(\delta_2-\sigma)\to N(0,2\sigma^4/(4\sigma^2))$$

So now we can find the asymptotic relative efficiency, $ARE$, of $\delta_1$ and $\delta_2.$

$$ARE=\frac{Var(\delta_2)}{Var(\delta_1)}=\frac{\sigma^2/2}{\sigma^2(\pi-2)/2}=\frac{1}{\pi-2}\approx0.87$$

This means I can use estimator $\delta_1$ with sample size $n_1$ or I could use estimator $\delta_2$ with sample size $n_2=ARE*n_1$ and obtain the same estimation accuracy (for large sample sizes).